Problem: What value of $x$ will give the minimum value for $x^2- 10x + 24$?
Answer: We start by completing the square.  \[x^2-10x+24=(x-5)^2-1.\] Since the square of a real number is at least 0, $(x-5)^2\ge 0$ and $(x-5)^2-1 \ge -1.$ Thus, the minimum value of the quadratic is $-1,$ which occurs when $x=\boxed{5}.$